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At $p\in D$, each of the given ambient rotations fixes $p$, rotates its tangent plane and preserves the local surface. Naturality of the shape operator under that rotation makes $S_p$ commute with every planar rotation. A symmetric real two-by-two operator commuting with all rotations is scalar, so every point of $D$ is an umbilical point. The trace-free part of the shape operator is continuous. Density therefore makes it zero everywhere on $S$.

Locally write $S=\kappa I$. The Codazzi equation gives $(X\kappa )Y=(Y\kappa )X$ for tangent fields $X,Y$; choosing independent fields proves $d\kappa =0$. Thus the surface is locally a plane if $\kappa =0$, or has constant spherical centersince its differential is $I-S/\kappa =0$. A change of normal reverses both $N$ and $\kappa$, leaving this center unchanged, so these local descriptions patch across the connected surface. In the flat case the whole surface lies in one plane, where a regular surface without boundary is open; a nonempty such set cannot also be compact and closed. Hence the flat case is excluded. In the other case $S$ lies on one sphere of radius $1/|\kappa |$. Its inclusion in that sphere is locally open, since both have dimension two, and its compactness makes the image closed. Connectedness of the sphere forces the image to be the entire sphere. Therefore $S$ is a sphere.

Let $H_k$ be the open hyperbolic half-plane beyond ${\gamma }_k$, opposite the central region $U$. These three half-planes have disjoint closures in the disk, and hyperbolic reflection $J_k$ exchanges $H_k$ and its complementary half-plane. In particular $J_k(U^{\circ })\subset H_k$ and $J_i(H_j)\subset H_i$ when $i\ne j$. A reduced nonempty reflection word therefore sends an interior point of $U$ into the half-plane corresponding to its first letter. It cannot be the identity. This ping-pong lemma argument proves that the reflection group isThe usual reflection tiling makes $U$ a fundamental domain for $W$: successive reflection across an offending side reduces a point to $U$, and the disjoint-side configuration yields locally finite tiles. Equivalently this is the disjoint-geodesic case of the reflection fundamental-domain theorem; there are no vertex relations because the sides never meet.

The orientation-preserving subgroup consists of even words. Every pair $J_i{J}_j$ can be expressed in $A=J_2{J}_1$ and $B=J_3{J}_2$: the other nontrivial pairs are their inverses and $J_3{J}_1=BA$, $J_1{J}_3=A^{-1}{B}^{-1}$. Thus it is exactly $G$. A fundamental set is $U\cup J_{2}U$, with one representative chosen on each paired side; its interior is a fundamental domain. To see there are no relations between $A,B$, the reflection-word tree has vertices the reduced words and edges joining $w$ to $w{J}_i$. The even subgroup acts freely on this tree; its quotient has two vertices joined by three edges. Choosing the $J_2$ edge as a spanning tree leaves two independent loops, represented by $A$ and $B$. Reduced loops in a graph form a free group, soThe quotient is the double of the central three-sided region along its geodesic sides: an orientable pair of pants, topologically a sphere with three points removed. Its three ends are hyperbolic funnels, not cusps, because the sides are ultraparallel and the peripheral transformations have positive translation lengths. The compact convex core has three geodesic boundary circles; the complete quotient itself has no boundary.

For a point $\mathbf{x}$ on the plane, the Cauchy-Schwarz inequality gives $|d|=|\mathbf{x}\cdot \mathbf{n}|\le |\mathbf{x}|$, because $\mathbf{n}$ is a unit vector. Equality is achieved at $\mathbf{x}=d\mathbf{n}$. Thus the distance from the origin is $|d|$.

Let $\delta =\mathbf{p}\cdot \mathbf{n}-d$ and let $\mathbf{q}=\mathbf{p}-\delta \mathbf{n}$ be the perpendicular projection of the sphere's center onto the plane. Every point of the plane has form $\mathbf{x}=\mathbf{q}+\mathbf{u}$ with $\mathbf{u}\cdot \mathbf{n}=0$. The Pythagorean theorem givesIf $|\delta |=r$, the sphere equation therefore forces $\mathbf{u}=0$. There is exactly one contact point, $\mathbf{q}$; it lies on both surfaces.

Write $\tau =\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})>0$. The nonzero scalar triple product proves that $\mathbf{a},\mathbf{b},\mathbf{c}$ are linearly independent, so they form a positively oriented basis of ${\mathbb{R}}^3$ and define a nondegenerate tetrahedron.

For the three faces through the origin, choose the inward unit vectorsTheir signs are correct because each has positive inner product with the opposite vertex vector. Write the center as $\mathbf{p}=\alpha \mathbf{a}+\beta \mathbf{b}+\gamma \mathbf{c}$. It lies on the interior side of every face. Tangency of the interior sphere therefore means signed distance $r$, not $-r$, from each of those three planes. Taking inner products givesSolving proves the center formulaThis is the sphere center from three tetrahedron face distances; the interior hypothesis selects these signs rather than an exterior center.

Finally set $\mathbf{N}=\mathbf{a}\times \mathbf{b}+\mathbf{b}\times \mathbf{c}+\mathbf{c}\times \mathbf{a}$. The scalar triple product identities give $\mathbf{N}\cdot \mathbf{a}=\mathbf{N}\cdot \mathbf{b}=\mathbf{N}\cdot \mathbf{c}=\tau$. In particular $\mathbf{N}\ne 0$, and the plane through the three vertices isThe second formula follows by normalizing $\mathbf{N}$ to a unit vector and applying the first distance result.

The quadratic expression is ${\mathbf{x}}^{T}A\mathbf{x}$. Normalize the preceding eigenvectors to obtain an orthonormal basisWith the orthogonal matrix $U=({\mathbf{u}}_1{\mathbf{u}}_2{\mathbf{u}}_3)$ and coordinates $\mathbf{y}=U^T\mathbf{x}$, the surface equation becomes $y_1^2+2y_2^2+4y_3^2=1$. All coefficients are positive, so it is an ellipsoid, with principal semiaxes $1,1/\sqrt{2},1/2$ along the respective eigenvectors. This is principal-axis reduction of a quadric.

A map to the unit sphere is $\mathbf{x}\mapsto \mathrm{diag}(1,\sqrt{2},2)U^T\mathbf{x}$. One explicit matrix isIndeed $B^{T}B=A$, so $|B\mathbf{x}{|}^2={\mathbf{x}}^{T}A\mathbf{x}$. Since $B$ is invertible, every point on the unit sphere has a preimage on the ellipsoid; the map is onto, not merely into. The requested $B$ is not unique, since an additional orthogonal matrix on the left preserves the unit sphere.

Choose a generator $[S^n]$ of ${\tilde{H}}_n(S^n;\mathbb{Z})\cong \mathbb{Z}$. The mapping degree of a continuous map is the integer determined byFor $n\ge 1$ this is ordinary top-dimensional homology; using reduced homology also gives a convention for $n=0$. Domain and target use the same orientation, so reversing the chosen generator in both leaves the integer unchanged. Homotopy invariance of homology makes the mapping degree invariant under homotopy.

We interpret the manifold criterion for a suspension in the category of nonempty topological manifolds without boundary, and “sphere” as homeomorphic to the standard sphere. Let $X$ have dimension $d$. If $X\cong S^d$, an explicit homeomorphism from its suspension to $S^{d+1}$ isIt is a continuous map that is a bijection: the two collapsed ends become the poles, and every other point determines $t$ and $x$ uniquely. The compact-to-Hausdorff continuous bijection theorem makes it a homeomorphism.

Conversely suppose $SX$ is a topological manifold. Its open part $X\times (0,1)$ shows its dimension is $d+1$. At a suspension vertex $v$ there is an open cone neighbourhood $U$ on $X$. It is contractible, while $U\setminus \{v\}$ admits a deformation retraction onto $X$. The long exact sequence in relative homology and excision give the local homology from a link calculationA point of a $(d+1)$-dimensional topological manifold without boundary has local homology $\mathbb{Z}$ in degree $d+1$ and zero in all other degrees. ConsequentlyThus $X$ is an integral homology sphere. This conclusion alone does not identify its homeomorphism type.

If $d\ge 2$, it is also simply connected. Choose a cone neighbourhood $U$ of the vertex and an open coordinate ball $B$ around the vertex inside $U$. By compactness of $X$, sufficiently short entire cone neighbourhoods fit inside any neighbourhood of $v$: the inverse image of that neighbourhood contains $X\times \{0\}$, and a finite subcover supplies one uniform collar length. Choose such a shorter cone $V\subset B$. The inclusions of punctured neighbourhoods factor asThe composite induces an isomorphism of fundamental groups, since both cone punctures are products of $X$ with intervals and their inclusion is a homotopy equivalence. The middle punctured ball has the homotopy type of $S^d$ and is simply connected for $d\ge 2$. Hence the composite is also zero on the fundamental group, forcing ${\pi }_1(X)=0$. This uses the nested neighbourhoods, not an unsupported claim that an arbitrary punctured neighbourhood is a sphere.

The Hurewicz theorem, applied successively to the vanishing lower homology groups, now gives ${\pi }_j(X)=0$ for $j<d$ and ${\pi }_d(X)\cong H_d(X)\cong \mathbb{Z}$. A representative of its generator is a continuous map $S^d\to X$ inducing an isomorphism in every integral homology degree. Topological manifolds have the homotopy type of CW complexes, so the homological Whitehead theorem makes it a homotopy equivalence. We have proved that $X$ is a homotopy sphere.

At this point the topological classification theorem is essential: the topological generalized Poincare theorem says that a closed homotopy sphere is homeomorphic to the standard sphere. Applying it completes the converse for $d\ge 2$. For $d=1$, the local calculation makes $X$ a connected compact one-dimensional manifold without boundary, hence a circle. For $d=0$, a compact zero-dimensional manifold is a finite set, and ${\tilde{H}}_0(X)\cong \mathbb{Z}$ forces exactly two points. Thus it is $S^0$. Altogether,The last classification step is not a consequence of homology alone, and we make no assertion of diffeomorphism with a prescribed smooth structure. If the word manifold allows boundary on both sides, the printed assertion needs this correction: the suspension of the interval is a closed disk. For instance $[x,t]\mapsto ((2x-1)(1-|2t-1|),2t-1)$ identifies it with the filled diamond $|u|+|v|\le 1$, although the interval is not a sphere.

To construct sphere maps of arbitrary integer degree, first use $h_m:S^1\to S^1$, $h_m(z)=z^m$ for any integer $m$, including the constant map when $m=0$. Lifting its argument to the real line gives $\theta \mapsto m\theta$, so the fundamental one-cycle is sent to $m$ times itself. Therefore $\mathrm{deg}{h}_m=m$.

Two cone neighbourhoods cover a suspension; their intersection admits a deformation retraction onto the original space. The reduced Mayer–Vietoris theorem gives a natural suspension isomorphismNaturality means that the suspension of a continuous map induces the same integer multiplier. This proves degree under suspension. Using the standard identification $S(S^k)\cong S^{k+1}$, setThus every integer occurs in every positive sphere dimension, with negative degrees handled just as well as positive ones.

If “every $n$” includes zero, that final printed assertion is false. The generator of ${\tilde{H}}_0(S^0)$ is the difference of its two points. A map of a two-point set is the identity, the transposition, or one of two constant maps, giving preciselyThese degrees of maps of the zero-sphere are the explicit exception to the arbitrary-degree construction.

For a CW complex $X$, let $X^k$ be its $k$-skeleton. The cellular chain complex isThe isomorphism uses excision and $H_k(D^k,S^{k-1})\cong \mathbb{Z}$, after choosing an orientation for each cell. In degree zero, take $X^{-1}=\varnothing$ and the free group on the vertices. The differential is the compositeThe long exact sequences of skeleton pairs imply $d_{k-1}{d}_k=0$. The cellular boundary formula describes its coefficient at a $(k-1)$-cell as the mapping degree of the attaching sphere after all other $(k-1)$-cells are collapsed. In degree one this is the signed difference of the two endpoints. The cellular homology theorem identifies the homology of this complex with singular homology: relative homology of successive skeletons is concentrated in their cell dimension, and the exact skeleton sequences leave precisely these kernels modulo images. This is the construction, rather than merely a count of cells.

The filtration ${\mathbb{CP}}^0\subset {\mathbb{CP}}^1\subset {\mathbb{CP}}^2$ gives one cell in each of dimensions zero, two and four, since ${\mathbb{CP}}^k\setminus {\mathbb{CP}}^{k-1}\cong {\mathbb{C}}^k$. There are no odd-dimensional cells, so every cellular differential vanishes. ThusThe complex projective line has one zero-cell and one two-cell. Its product with itself has one zero-cell, two two-cells and one four-cell. Its cellular chain complex again has zero differentials, givingA homeomorphism induces isomorphisms on homology groups. The second groups have different ranks, and therefore

Now orient the Complex projective plane by its complex coordinates. The inclusion of its two-skeleton ${\mathbb{CP}}^1$ identifies the fundamental class of a projective line with a generator $h$ of $H_2({\mathbb{CP}}^2;\mathbb{Z})\cong \mathbb{Z}$: the cellular generator has no incoming three-cell boundary and no outgoing boundary.

Take the projective lines $L_0=\{z_2=0\}$ and $L_1=\{z_1=0\}$. A path of unitary coordinate transformations takes one to the other, so their oriented homology classes are both $h$. They meet in the single point $[1:0:0]$. In its affine chart, let $w=z_1/z_0$ and $v=z_2/z_0$. Their tangent spaces are the complex $w$-axis and complex $v$-axis, so the intersection is a transverse intersection. The ordered real basesconcatenate to the complex orientation of the ambient four-manifold. Thus the local sign is $+1$, andThis computes the intersection form using distinct representatives of the same class, avoiding an ill-defined attempt to count a line intersecting itself as a set. Bilinearity now givesEquivalently, the map $H_2\to \mathrm{Hom}(H_2,\mathbb{Z})$ sending $a$ to $\lambda (a,-)$ is an isomorphism. This is exactly a unimodular intersection pairing, and proves projective lines generate a unimodular intersection form. Reversing the ambient orientation would change the matrix to $(-1)$ and would still be unimodular.

The multiplicative norm first rules out zero divisors: if $xy=0$ and $x\ne 0$, then $0=|x||y|$ forces $y=0$. If the dimension is zero the claimed bound already holds, so suppose it is positive.

It is worth checking that the printed description of a ring with an underlying real vector space really permits differentiation of multiplication. This follows from the norm identity even if compatibility with real scalars is not separately stated. For fixed $x$, the map $L_x(y)=xy$ is additive by distributivity, andIt is therefore continuous. Additivity gives rational linearity. Approximating a real number by rationals and using continuity proves $L_x(ty)=t{L}_x(y)$ for every real $t$. The same argument holds in the other variable. This is automatic real bilinearity from a multiplicative norm, so multiplication is a bilinear map between finite-dimensional vector spaces, and hence is smooth.

Let $S=\{x\in A:|x|=1\}$ be the unit sphere of the given positive-definite bilinear form. It is a copy of $S^{n-1}$. The squaring mapis well defined because $|x^2|=|x{|}^2=1$, and is smooth. Commutativity gives its derivativeFor $x\in S$ and $v\ne 0$, the norm identity gives $|D{Q}_x(v)|=2|v|$, so the derivative is injective, including on the tangent space $T_{x}S=x^{\perp }$. It takes tangent vectors into $T_{x^2}S$ since $Q$ maps the sphere to itself. Alternatively, polarization of $|xy{|}^2=|x{|}^2|y{|}^2$ gives $⟨xy,xz⟩=|x{|}^2⟨y,z⟩$, so $⟨x^2,2xv⟩=2⟨x,v⟩=0$ directly.

The only identifications under squaring are antipodal:The factorization uses commutativity, and the last implication uses the absence of zero divisors. Conversely $Q(-x)=Q(x)$. It follows that $Q$ induces an injective smooth mapThe antipodal quotient map is a local diffeomorphism, so the injectivity of $DQ$ implies injectivity of $Dq$. Since Real projective space is compact, the stated compact injective immersion is an embedding theorem makes $q$ a closed embedding.

For $n\ge 2$ its source and target have the same dimension $n-1$. The inverse function theorem makes its image open as well as closed in the connected sphere $S^{n-1}$. It is nonempty, so the image is the whole sphere; consequently $q$ is a diffeomorphism. But for $n\ge 3$ this is impossible:Indeed, the antipodal two-sheeted covering map $S^{n-1}\to {\mathbb{RP}}^{n-1}$ has simply connected total space in this range, and is its universal cover, with deck group $\mathbb{Z}/2$. A sphere of dimension at least two is simply connected. These fundamental groups cannot be those of homeomorphic spaces.

We have obtained the dimension bound for a commutative Euclidean normed algebra:The real numbers and complex numbers with their usual products and Euclidean norms attain dimensions one and two. In dimension two, the induced map ${\mathbb{RP}}^1\to S^1$ is entirely compatible with the argument: the circle squaring map identifies exactly the two antipodal preimages. We neither assume a multiplicative identity nor use a classification of real division algebras; the squaring argument proves the bound from the supplied hypotheses.